Integrand size = 77, antiderivative size = 25 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=x+e^{\frac {e^{-4 x^2} x}{\log (x)}} (4-5 (7+x)) \]
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\[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (1-5 e^{\frac {e^{-4 x^2} x}{\log (x)}}+\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x) \left (1-\log (x)+8 x^2 \log (x)\right )}{\log ^2(x)}\right ) \, dx \\ & = x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x) \left (1-\log (x)+8 x^2 \log (x)\right )}{\log ^2(x)} \, dx \\ & = x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \left (\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x)}{\log ^2(x)}+\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} \left (-31-5 x+248 x^2+40 x^3\right )}{\log (x)}\right ) \, dx \\ & = x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x)}{\log ^2(x)} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} \left (-31-5 x+248 x^2+40 x^3\right )}{\log (x)} \, dx \\ & = x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \left (\frac {31 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log ^2(x)}+\frac {5 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {31 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log (x)}-\frac {5 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log (x)}+\frac {248 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^2}{\log (x)}+\frac {40 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^3}{\log (x)}\right ) \, dx \\ & = x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+5 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log ^2(x)} \, dx-5 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log (x)} \, dx+31 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log ^2(x)} \, dx-31 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log (x)} \, dx+40 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^3}{\log (x)} \, dx+248 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^2}{\log (x)} \, dx \\ \end{align*}
Time = 5.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=e^{\frac {e^{-4 x^2} x}{\log (x)}} (-31-5 x)+x \]
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Time = 0.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x +{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}} \left (-31-5 x \right )\) | \(22\) |
parallelrisch | \(-5 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}} x -31 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}}+x\) | \(38\) |
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=-{\left (5 \, x + 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )} + x \]
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Time = 57.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=x + \left (- 5 x - 31\right ) e^{\frac {x e^{- 4 x^{2}}}{\log {\left (x \right )}}} \]
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\[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int { \frac {{\left (e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (5 \, e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (40 \, x^{3} + 248 \, x^{2} - 5 \, x - 31\right )} \log \left (x\right ) - 5 \, x - 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )}\right )} e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )^{2}} \,d x } \]
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\[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int { \frac {{\left (e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (5 \, e^{\left (4 \, x^{2}\right )} \log \left (x\right )^{2} - {\left (40 \, x^{3} + 248 \, x^{2} - 5 \, x - 31\right )} \log \left (x\right ) - 5 \, x - 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )}\right )} e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )^{2}} \,d x } \]
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Timed out. \[ \int \frac {e^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{-4\,x^2}\,\left ({\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-4\,x^2}}{\ln \left (x\right )}}\,\left (-5\,{\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+\left (40\,x^3+248\,x^2-5\,x-31\right )\,\ln \left (x\right )+5\,x+31\right )\right )}{{\ln \left (x\right )}^2} \,d x \]
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